Đáp án:
a) $\left[\begin{array}{l}x =\dfrac{\pi}{6} + k2\pi\\x =\dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x =\arctan\dfrac14 + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
a) $\sin x =\dfrac12$
$\to \sin x =\sin\dfrac{\pi}{6}$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{6} + k2\pi\\x =\dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $4\tan^2x - 5\tan x +1 = 0\qquad (*)$
$ĐK:\, x \ne \dfrac{\pi}{2} +n\pi$
$(*)\Leftrightarrow (\tan x -1)(4\tan x -1)=0$
$\to \left[\begin{array}{l}\tan x = 1\\\tan x =\dfrac14\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x =\arctan\dfrac14 + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
